electric field of a line charge and point charge


by GravityGirl
Tags: charge, electric, field, line, point
GravityGirl
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#1
May23-07, 02:39 PM
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A thin rod 39.2 cm long is charged uniformly with a positive charge density of 72.0 mC/m. The rod is placed along the y-axis and is centered at the origin. A charge of +46.5 mC is placed 68.4 cm from the midpoint of the rod on the positive x-axis. Calculate the electric field at a point on the x-axis, which is halfway between the point charge and the center of the rod. (Express your answer in terms of the unit vector x. For example, if the electric field is -10.5x N/C, then input -10.5 N/C.)


ok..so i drew a picture and i see that the E field of te line charge will be in the x direction becuase that i perpendicular to the line

i also see that the point charge will be in the x direction as well becuase it is located on the x-axis


so since both charges are postive and the point where i want to know the E field at is between them i am going to subtract their individual E fields at that point to get the net

E=Eline-Epoint

Eline=2K(lamda)/r where r= half the distance betwen the line and point charge

Epoint=kq/r^2

so i have E=(2k(lamda)/r)-(kq/r^2)

does this look right?....if not, please tell me what i am doing wrong
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Doc Al
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#2
May23-07, 02:57 PM
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Quote Quote by GravityGirl View Post
ok..so i drew a picture and i see that the E field of te line charge will be in the x direction becuase that i perpendicular to the line
OK.

i also see that the point charge will be in the x direction as well becuase it is located on the x-axis
The field from that charge will point along the negative x-axis.


so since both charges are postive and the point where i want to know the E field at is between them i am going to subtract their individual E fields at that point to get the net

E=Eline-Epoint
OK. But I would prefer that you viewed it as adding the positive field from the line charge to the negative field from the point charge.

Eline=2K(lamda)/r where r= half the distance betwen the line and point charge
That formula applies to an infinitely long line charge. For a finite line charge you need to integrate.

Epoint=kq/r^2
OK.
neutrino
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#3
May23-07, 03:06 PM
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Quote Quote by Doc Al View Post
That formula applies to an infinitely long line charge. For a finite line charge you need to integrate.
Oops. I must've misread what GravityGirl wrote.

GravityGirl
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#4
May23-07, 03:18 PM
P: 29

electric field of a line charge and point charge


ok so if i am to integrat across the line charge then i would have

integral of Kdq/r^2=kQ/x(x-L) where L= the lenght of the rod and x= the distance to the point where i want to find the E field

so now i have (KQ/x(x-L))-(kQpoint/r^2)=E



but doesn't gauss's law also work (what i did above)
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#5
May23-07, 03:29 PM
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Quote Quote by GravityGirl View Post
ok so if i am to integrat across the line charge then i would have

integral of Kdq/r^2=kQ/x(x-L) where L= the lenght of the rod and x= the distance to the point where i want to find the E field
How did you get that answer? Doesn't look right to me. (What if x = L?)
but doesn't gauss's law also work (what i did above)
Gauss's law is only useful when there is enough symmetry to simplify the calculation of the electric flux. That's true with an infinite line of charge, since you know the field must be radial; but not true with a finite line of charge.
GravityGirl
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#6
May23-07, 03:31 PM
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i got the simplified answer to the integral from my book....but what about if is had ....

=k/r^2(intergral of dq)=kQ/r^2

does that look better??
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#7
May23-07, 03:39 PM
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Quote Quote by GravityGirl View Post
i got the simplified answer to the integral from my book....but what about if is had ....

=k/r^2(intergral of dq)=kQ/r^2

does that look better??
That's only true for a point charge at one location. In a line charge, each charge element is at a different distance and direction. Read this: Electric Field of Line Charge
GravityGirl
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#8
May23-07, 03:46 PM
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i see that the equation Ez is the same equation that i wa using in the firs attempt
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#9
May23-07, 03:46 PM
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so should Eline=2k(lamda)/r?
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#10
May23-07, 06:17 PM
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Quote Quote by GravityGirl View Post
i see that the equation Ez is the same equation that i wa using in the firs attempt
No it isn't. Make sure you are looking at the right equation.

Quote Quote by GravityGirl View Post
so should Eline=2k(lamda)/r?
No, as I said before, that only applies to an infinitely long line of charge.
sonutulsiani
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#11
Feb14-10, 05:52 PM
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Ok here can't we just take it as a distance of 68.4/2 = 34.1 cm from the midpoint of the rod? And then just integrate k dq/r^2? with r = 34.1 cm which can be converted to metres later.
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#12
Feb14-10, 05:59 PM
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Quote Quote by sonutulsiani View Post
Ok here can't we just take it as a distance of 68.4/2 = 34.1 cm from the midpoint of the rod? And then just integrate k dq/r^2? with r = 34.1 cm which can be converted to metres later.
You cannot treat the line charge as if its charge were concentrated at its center, if that's what you're thinking.
sonutulsiani
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#13
Feb14-10, 07:47 PM
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No I am not thinking that

I am thinking that if we take the symmetry from the line charge from both the sides, then we get the x component of the E field at 34.1 cm right?
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#14
Feb15-10, 07:21 AM
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Quote Quote by sonutulsiani View Post
I am thinking that if we take the symmetry from the line charge from both the sides, then we get the x component of the E field at 34.1 cm right?
I still don't understand what you mean. To calculate the field from that line charge you must integrate the field from each element of charge along the line. By symmetry, you know that there will only be an x component of the field.
Dosmascerveza
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#15
Mar1-10, 08:26 PM
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you can integrate the function dE=kdQ/(r^2). dQ=lambda dy where lambda is the linear charge density Q/a where a is the length of the line of charge. and you find the y component by multiplying by the unit vector sin(phi) where phi is the angle between the x axis and r where r is the distance from P(x,y) to the upper limit of your line of charge. r=sqrt(x^2+y^2)


SO YOU GET...


S dE_y= kQ/(a) S[(y)r^(-3/2)]dy.

What you will get if you integrate the right side from 0 to a is the y component of your Electric field. To find the x component uses a similar process but the integral is a bit trickier (trig substitution) Then you can plug and chug away.

there is probably an easier way to do this but i think the long way helps build an intuition because now its clear as day for me.


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